Question

The square of the distance of the point \(\left( \frac{15}{7}, \frac{32}{7}, 7 \right)\) from the line \(\frac{x+1}{3} = \frac{y+3}{5} = \frac{z+5}{7}\) in the direction of the vector \(\mathbf{i} + 4\mathbf{j} + 7\mathbf{k}\) is:

• A. \(41\)
• B. \(44\)
• C. \(54\)
• D. \(66\)

Hint:

To find the square of the distance between two points, use the distance formula and then square the result.

Answer 1

The Correct Option is D

Step 1: Equation of the line and point. The line is given by: \[ L: \frac{x+1}{3} = \frac{y+3}{5} = \frac{z+5}{7} \] The point \( P = \left( \frac{15}{7}, \frac{32}{7}, 7 \right) \). Step 2: Find the coordinates of point Q. Let \( Q \) be the point on the line \( L \), and assume the parametric coordinates for \( Q \) are: \[ Q: \left( \lambda + \frac{15}{7}, 4\lambda + \frac{32}{7}, 7\lambda + 7 \right) \] Step 3: Use the condition that point \( Q \) lies on the line \( L \). The coordinates of \( Q \) should satisfy the line equation. Using the first equation \( \frac{x+1}{3} = \frac{y+3}{5} = \frac{z+5}{7} \), we find: \[ \frac{\lambda + \frac{15}{7} + 1}{3} = \frac{4\lambda + \frac{32}{7} + 3}{5} = \frac{7\lambda + 7 + 5}{7} \] \[ \Rightarrow 7\lambda + 22 = 21\lambda + 36 \] \[ \Rightarrow \lambda = -1 \] Step 4: Coordinates of point \( Q \). Substitute \( \lambda = -1 \) into the parametric form of \( Q \): \[ Q = \left( \frac{8}{7}, 4, 0 \right) \] Step 5: Find the distance \( PQ \). The distance between points \( P \left( \frac{15}{7}, \frac{32}{7}, 7 \right) \) and \( Q \left( \frac{8}{7}, 4, 0 \right) \) is given by: \[ PQ = \sqrt{\left( \frac{15}{7} - \frac{8}{7} \right)^2 + \left( \frac{32}{7} - 4 \right)^2 + (7 - 0)^2} \] \[ PQ = \sqrt{\left( \frac{7}{7} \right)^2 + \left( \frac{32}{7} - \frac{28}{7} \right)^2 + 7^2} \] \[ PQ = \sqrt{1^2 + \left( \frac{4}{7} \right)^2 + 49} \] \[ PQ = \sqrt{1 + \frac{16}{49} + 49} = \sqrt{\frac{65}{49} + 49} \] \[ PQ = \sqrt{\frac{65 + 2401}{49}} = \sqrt{\frac{2466}{49}} = \sqrt{66} \] \[ \Rightarrow PQ^2 = 66 \]